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It is observed that 2/7 f the balls that have red color also have

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It is observed that 2/7 f the balls that have red color also have [#permalink]

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New post 28 Jan 2018, 05:03
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Question Stats:

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There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f the balls that have red color also have green color while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT
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It is observed that 2/7 f the balls that have red color also have [#permalink]

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New post Updated on: 03 Feb 2018, 22:34
1
ven diagram..
87=R+G-Both RG

given that ...Both = 2R/7 = 3G/7; we need to find Both/total

87= 7Both/2 + 7Both/3 - Both

Both = 18

therefore , 18/87 = 6/29 D


Corrected G in 3G/7
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Originally posted by doomedcat on 28 Jan 2018, 05:29.
Last edited by doomedcat on 03 Feb 2018, 22:34, edited 1 time in total.
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Re: It is observed that 2/7 f the balls that have red color also have [#permalink]

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New post 02 Feb 2018, 09:45
Still not able to understand. Could someone explain ?
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It is observed that 2/7 f the balls that have red color also have [#permalink]

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New post 03 Feb 2018, 22:32
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poul_249 wrote:
Still not able to understand. Could someone explain ?


You have to visualise the problem as set problem...
Given 87 Balls with Red and Green colour
So the set equation becomes; Total (Red & Green Balls) = #Red Balls + # Green Balls - Both (Red & Green)

Given is 2/7 of Red balls also are Green ==> Both ( Red and green) = \(\frac{2}{7}\) * # Red Balls
Also it is stated that 3/5 of Green balls are red, which is again stating that ==> Both ( Red and green) = \(\frac{3}{7}\) * # Green Balls

Essentially it means that #Red Balls = \(\frac{7}{2}\)* Both (Red and Green)
Also it means that #Green Balls = \(\frac{7}{3}\)* Both (Red and Green)

Now coming back to original equation Total (Red & Green Balls) = #Red Balls + # Green Balls - Both (Red & Green)

Substituting the values we got above

87= \(\frac{7}{2}*Both + \frac{7}{3}*Both - Both\)
So Both = 18

We are asked to find Both/Total =\(\frac{18}{87}\) = \(6/29\)

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Re: It is observed that 2/7 f the balls that have red color also have [#permalink]

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New post 15 Feb 2018, 22:07
selim wrote:
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f the balls that have red color also have green color while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT



R Red
G Green
X Both

Given:
Total Balls = 87
(2/7)*R = X ----> R = (7/2)*X ---->(1)
(3/7)*G = X ----> G = (7/3)*X ---->(2)

formula : T = R+G - both(X)
Applying 1 and 2 in formula
87= (7/2)*X + (7/3)*X - X
Solving, we get X= 18

Now fr fraction X/T = 18/87 = 6/29

Answer : D 6/29
Re: It is observed that 2/7 f the balls that have red color also have   [#permalink] 15 Feb 2018, 22:07
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It is observed that 2/7 f the balls that have red color also have

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